I was messing around with some formulas for the Polylogarithm $\operatorname{Li}_s(z)$ when I got the result that $0 = \frac{\sqrt{\pi}}{2}$ which is clearly absurd. Here are my calculations:
Under the subsection "Series representations" on the Wikipedia page over Polylogarithms the following result is listed:
$$\operatorname {Li} _{s}(e^{\mu })=\Gamma (1-s)\,(-\mu )^{s-1}+\sum _{k=0}^{\infty }{\zeta (s-k) \over k!}\,\mu ^{k}\,.$$
This result holds for $|\mu| < 2π$ and, thanks to the analytic continuation provided by the zeta functions, for all s ≠ 1, 2, 3, ...
Plugging in $\mu = -1$ and $s = -\frac{1}{2}$ we get:
$$\operatorname {Li} _{-\frac{1}{2}}\left(\frac{1}{e}\right)=\Gamma \left(\frac{3}{2}\right)+\sum _{k=0}^{\infty }{\zeta\left(-\frac{1}{2}-k\right) \over k!}(-1)^{k}\,$$
Then expanding $\zeta\left(-\frac{1}{2}-k\right)$ we get:
\begin{align} \Gamma \left(\frac{3}{2}\right)+\sum _{k=0}^{\infty }{\zeta\left(-\frac{1}{2}-k\right) \over k!}(-1)^{k} & = \Gamma \left(\frac{3}{2}\right)+\sum _{k=0}^{\infty } {(-1)^{k}\over k!}\sum _{n=1}^{\infty} n^{\frac{1}{2}+k} \\\\ & = \Gamma \left(\frac{3}{2}\right)+\sum _{k=0}^{\infty}\sum _{n=1}^{\infty} {(-1)^{k}\over k!} n^{\frac{1}{2}+k} \\\\ &= \Gamma \left(\frac{3}{2}\right)+\sum _{n=1}^{\infty}\sum _{k=0}^{\infty} {(-1)^{k}\over k!} n^{\frac{1}{2}+k} \\\\ & = \Gamma \left(\frac{3}{2}\right)+\sum _{n=1}^{\infty}\sqrt{n}\sum _{k=0}^{\infty} {(-1)^{k}\over k!} n^{k} \\\\ & = \Gamma \left(\frac{3}{2}\right)+\sum _{n=1}^{\infty}\frac{\sqrt{n}}{e^n}\end{align}
But since:
$$\sum _{n=1}^{\infty}\frac{\sqrt{n}}{e^n} = \operatorname {Li} _{-\frac{1}{2}}\left(\frac{1}{e}\right)$$
We get the absurd result that:
$$\operatorname {Li} _{-\frac{1}{2}}\left(\frac{1}{e}\right) = \frac{\sqrt{\pi}}{2}+\operatorname {Li} _{-\frac{1}{2}}\left(\frac{1}{e}\right)$$
Which clearly isn't true. Where was I wrong? Was it the part where I exchanged the sums?